Question: Complete the square to solve for $x$. $4x^{2}+4x-3 = 0$
First, divide the polynomial by $4$ , the coefficient of the $x^2$ term. $x^2 +x - \dfrac{3}{4} = 0$ Move the constant term to the right side of the equation. $x^2 +x = \dfrac{3}{4}$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $1$ , so half of it would be $\dfrac{1}{2}$ , and squaring it gives us ${\dfrac{1}{4}}$ $x^2 +x { + \dfrac{1}{4}} = \dfrac{3}{4} { + \dfrac{1}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x + \dfrac{1}{2} )^2 = 1$ Take the square root of both sides. $x + \dfrac{1}{2} = \pm1$ Isolate $x$ to find the solution(s). $x = -\dfrac{1}{2}\pm1$ The solutions are: $x = \dfrac{1}{2} \text{ or } x = -\dfrac{3}{2}$ We already found the completed square: $( x + \dfrac{1}{2} )^2 = 1$